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Proof step 27 from proof subarray-eq-elim (previous step)

# ar = m + n, ar = subarray(ar, 0, n) + subarray(ar, n, m) ⊦ ∃ ar0, ar1. ar = ar0 + subarray(ar, # ar0, m) + ar1 ∧ # ar0 = n

User applied rule exists right

And obtained the following new goals

Proof step 28:
⊦ ar + mkarray(0) = ar
Proof step 29:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1

Rule argument

pos: right 1 terms: subarray(ar, 0, n), mkarray(0) abort: F discard: T computed: F

Additional information

Used simplifier rules: append-zero-right: ⊦ ar + mkarray(0) = ar;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ # subarray(ar, m, n) = n; (s) Used terms: subarray(ar, 0, n), mkarray(0)